Simplify the following expression: $y = \dfrac{-8x^2+5x+3}{x - 1}$
Explanation: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-8)}{(3)} &=& -24 \\ {a} + {b} &=& &=& {5} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-24$ and add them together. Remember, since $-24$ is negative, one of the factors must be negative. The factors that add up to ${5}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-3}$ and ${b}$ is ${8}$ $ \begin{eqnarray} {ab} &=& ({-3})({8}) &=& -24 \\ {a} + {b} &=& {-3} + {8} &=& 5 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-8}x^2 {-3}x) + ({8}x +{3}) $ Factor out the common factors: $ x(-8x - 3) - 1(-8x - 3)$ Now factor out $(-8x - 3)$ $ (-8x - 3)(x - 1)$ The original expression can therefore be written: $ \dfrac{(-8x - 3)(x - 1)}{x - 1}$ We are dividing by $x - 1$ , so $x - 1 \neq 0$ Therefore, $x \neq 1$ This leaves us with $-8x - 3; x \neq 1$.